Question: The lifespans of tigers in a particular zoo are normally distributed. The average tiger lives $26.4$ years; the standard deviation is $4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a tiger living between $14.4$ and $34.4$ years.
Solution: $26.4$ $22.4$ $30.4$ $18.4$ $34.4$ $14.4$ $38.4$ $99.7\%$ $95\%$ $2.35\%$ $2.35\%$ We know the lifespans are normally distributed with an average lifespan of $26.4$ years. We know the standard deviation is $4$ years, so one standard deviation below the mean is $22.4$ years and one standard deviation above the mean is $30.4$ years. Two standard deviations below the mean is $18.4$ years and two standard deviations above the mean is $34.4$ years. Three standard deviations below the mean is $14.4$ years and three standard deviations above the mean is $38.4$ years. We are interested in the probability of a tiger living between $14.4$ and $34.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the tigers will have lifespans within 3 standard deviations of the average lifespan. It also tells us that $95\%$ of the tigers will have lifespans within 2 standard deviations of the mean. That leaves $99.7\% - 95\% = 4.7\%$ of tigers between 2 and 3 standard deviations of the mean, or $2.35\%$ on either side of the distribution. The probability of a particular tiger living between $14.4$ and $34.4$ years is $\color{orange}{2.35\%} + {95\%}$, or $97.35\%$.